Question

A 20 g ball is fired horizontally with speed v0 toward a 100 g ball hanging motionless from a 1.0-m-long string. The balls un-dergo a head-on, perfectly elastic collision, after which the 100 g ball swings out to a maximum angle umax=50. What was v0?

Answer 1

Answer:7.93 m/s

Explanation:

Given

mass of ball $m_1=20\ gm$

Mass of hanging ball $m_2=100\ gm$

Length of string $L=1\ m$

Maximum angle turned $\theta _{max}=50^{\circ}$

$v_o$ is the initial velocity of ball 1  and 0 is the initial  velocity of ball 2

For Perfectly elastic final velocity of ball 1 and 2 is given by

$v_2'=\frac{2m_1}{m_1+m_2}\cdot v_1-\frac{m_1-m_2}{m_1+m_2}\cdot v_2$

$v_1'=\frac{m_1-m_2}{m_1+m_2}\cdot v_1+\frac{2m_2}{m_1+m_2}\cdot v_2$

where $v_1$and $v_2$ are the velocity of 1 and 2 before collision

thus $v_2'=\frac{2\times 20}{120}v_0-0$

$v_2'=\frac{v_o}{3}$

$v_1'=\frac{20-100}{120}\times v_o+0$

$v_1'=-\frac{2}{3}v_o$

By energy conservation on second ball we get

Kinetic energy=Potential Energy

$\frac{1}{2}m_2(v_2')^2=m_2gL(1-\cos \theta )$

$v_2'=\sqrt{2gl(1-\cos \theta )}$

$v_2'=\sqrt{2\times 9.8\times 1(1-0.642)}$

$v_2'=2.64\ m/s$

thus $v_o=3\times v_2'=7.93\ m/s$